已知p=x,yy=mq=想,yyax 1
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(x-y)(x+y)(xx-yy)=(x^2-y^2)(x^2-y^2)=x^4-2x^2y^2+y^4
X:Y:Z=1:2:3因为:14(XX+YY+ZZ)=(X+2Y+3Z)^214(XX+YY+ZZ)-(X+2Y+3Z)^2=013X^2+10Y^2+5Z^2-4XY-6XZ-12YZ=0(4X^2
(x-1)^2+(y-1)^2=1(1)当b=1时,则M点在圆上,与圆上的两点组成直角三角形,则说明PQ为直径,即直线过圆点(1,1).把圆点代入直线方程得:k=1==!第二小题有点小难.
圆C:x^2+y^2-2x-2y+1=0,即:(x-1)^2+(y-1)^2=1,圆心位于(1,1)将直线l:y=kx代入得:(k^2+1)x^2-2(k+1)x+1=0有两个根,要求△=4(k+1)
MP斜率为(1-y1)/(-x1),MQ斜率为(1-y2)/(-x2)∵MP⊥MQ,∴(1-y1)/(-x1)*(1-y2)/(-x2)=-11-y1-y2+y1y2=-x1x2式子1将y=kx代入圆
1、MP斜率为(1-y1)/(-x1),MQ斜率为(1-y2)/(-x2)∵MP⊥MQ,∴(1-y1)/(-x1)*(1-y2)/(-x2)=-11-y1-y2+y1y2=-x1x2式子1将y=kx代
(1)当b=1时,点M在圆上,又MP垂直MQ,所以直线l经过圆心.将圆心坐标(1,1)代人L的方程,得:k=1(2)
圆C:(x-1)^2+(y-1)^2=1(1)b=1,则M(0,1)在圆上,由MP垂直MQ可得直线L必过原点(直径所对圆周角为直角)k=1/1=1(2)计算写不下,说说思路,设P和Q点坐标,利用向量点
2/9再问:过程,谢谢再答:由题目得y/x=2/3xy/xx+yy-yy/xx-yy=y/x-(y/x)²=2/3-4/9=2/9
由P(4,-9),Q(-2,3)可得线段PQ的斜率k=(3+9)/(-2-4)=-2设M的坐标为(0,b),则线段PQ的方程为y=-2x+b把P(4,-9)代入方程可得-9=-8+b即b=-1∴y轴与
园C:x^2-2x+1+y^2-2y+1-2=0,即:(x-1)^2+(y-1)^2=1,圆心为(1,1),半斤为1.---与此题无关.第一问:点M(0,B),且MP⊥MQ,可得直线MP和直线MQ的斜
xx+yy+4x-6y+13=0整理得:(x+2)^2+(y-3)^2=0那么只有(x+2)=0(y-3)=0x=-2y=3(x^2-2x)/(x^2+3y^2)=(4+4)/(4+3*9)=8/31
x^2+x-x^2-y=3x-y=3(x-y)^2=9x^2+y^2-2xy=9(x^2+y^2)/2-xy=9/2
1、MP斜率为(1-y1)/(-x1),MQ斜率为(1-y2)/(-x2)∵MP⊥MQ,∴(1-y1)/(-x1)*(1-y2)/(-x2)=-11-y1-y2+y1y2=-x1x2式子1将y=kx代
(x+y)(x+y)=25x^2+2xy+y^2=25……(1)(x-y)(x-y)=9x^2-2xy+y^2=9……(2)(1)+(2)得:2x^2+2y^2=34x^2+y^2=17(1)-(2)
x^2-2x+y^2+6y+10=0(x-1)^2+(y+3)^2=0所以x=1,y=-3x+y=-2x^2表示x的平方
XX+YY+4X-6Y+13=0(X+2)²+(Y-3)²=0X+2=0Y-3=0X=-2Y=3X的Y次方=-8